We are given that $X$ is $Uniform[4,10]$ with mean $7$ and variance $3$.
We are first asked to figure out $P(X \leq 5 \text { or } X \geq 9$)
Well this is a uniform graph with height = $1/(10-4)=1/6$, and we can do the following:
$P(X \leq 5 \text { or } X \geq 9) = P(X \leq 5) + P(X\geq 9)=(5-4)/6+(10-9)/6=2/6=1/3$
Now I am asking the following:
Let $X_1, X_2, X_3, \dots, X_{16}$ be a random sample from this uniform distribution. We are asked to find Chebyshev's upper bound for $P(\overline{X} \leq 5 \text { or } \overline{X} \geq 9)$ where $\overline{X}=\dfrac{\sum_{i=1}^{16}X_i}{16}$
Again, we can break up the probability to be as follows:
$P(\overline{X} \leq 5) + P(\overline{X} \geq 9)\leq \text{upper bound}$
Chebyshev Inequality is the following: $P(|X-\mu_x| \geq a)\leq \dfrac{var(x)}{a^2}$
So, for $P(\overline{X}\leq 5)$, we have $P(|\overline{X}-7|\leq5)=1-\dfrac{3}{25}$
For $P(\overline{X}\geq 9)$, we have $P(|\overline{X}-7|\geq 9)=\dfrac{3}{9^2}$
So in total, the upper bound must be $1-\dfrac{3}{25}+\dfrac{3}{9^2}=0.917$
Is this correct? I don't see where we used $\overline{X}$ in this. There would be nothing different if I considered $X$, instead of $\overline{X}$, so I think I am missing something.
Maybe the $var(x)$ should actually be $\sigma^2/n=3/16$. But in my notes, I have that this is only the case if $X_1, X_2, \dots, X_n$ are $N(\mu, \sigma^2)$. Here we have that they are uniform, so I am not sure.
Hint:
$P(X \leq 5 \text { or } X \geq 9) = P(|X-7| \geq 2) $
two away from mean on both sides..